Chapter 2
Polynomial, Power, and
Rational Functions
Mr. J. Focht
Pre-Calculus
OHHS
2.7 Solving Equations
in One Variable
 Solving
Rational Equations
 Extraneous Solutions
 Applications
What You’ll Do

Stewart Cannery will package tomato juice in
2-liter cylindrical cans. Find the radius and
height of the cans if the cans have a surface
area of 1000 cm2.
Solving Rational Equations
Multiply by x to
clear the
fraction
3x
x x   4x
x
x  3  4x
2
x2  4 x  3  0
( x  1)( x  3)  0
x 1  0
x3  0
x = 1 or x = 3
You may want to use the Quadratic Formula
Solving Rational Equations


Check your answers
x = 1 or x = 3
3
1  4
1
3
x 4
x
3
3  4
3
Now You Try
x2 x5 1


3
3
3
-1
Solving Rational Equations
2x
1
2

 2
x 1 x  3 x  4x  3
(x-1)(x-3)
2x
1 (x-1)(x-3)
2 (x-1)(x-3)


x 1
x 3
( x  1)( x  3)
(x-3)2x + x-1 = 2
Solving Rational Equations
(x-3)2x + x-1 = 2
2x2 – 6x + x-1 = 2
2
2x – 5x - 3 = 0
(2x + 1)(x – 3) = 0
x = -½ or x = 3
Check the Answers
2x
1
2

 2
x 1 x  3 x  4x  3
2   12 
x = -½ or x = 3

1

   1     3   
1
2
1
2
8

21
1 2
2
8
21
2
 4   12   3
Check the Answers
2x
1
2

 2
x 1 x  3 x  4x  3
x = -½ or x = 3
23
1
2

 2
3 1 3  3 3  4  3  3
3 doesn’t work. It is an extraneous root.
Check Graphically
Doesn’t
cross
the xaxis at 3
Crosses
the xaxis at
0.5
[-5,5] x [-10, 10]
Now You Try
Solve algebraically. Confirm graphically.
3x
1
7

 2
x  5 x  2 x  3x  10
1
- ,2
3
Now You Try
3
6
3 x
 2

x  2 x  2x
x
-2, 0
Mixture Problems

How much pure acid must
be added to 50mL of a 35%
acid solution to produce a
mixture that is 75% acid?
x mL
mL of pure acid
 concentrat ion of acid
mL of mixture
Pure acid = x + 35%(50) = x + 17.5
Mixture = x + 50
Mixture Problem
mL of pure acid
 concentrat ion of acid
mL of mixture
x + 17
 0.75
x + 50
x  17  0.75(x+50)
x  17  0.75x + 37.5
0.25 x  20.5
x  82
Mixture Problems

Verify graphically
x + 17
 0.75
x + 50
[0, 160] x [-1, 1]
Writing to Learn

You would add 82 mL of
pure acid to the 50 mL of
35% solution to create a
75% solution.
Now You Try
Suppose that x mL of pure acid are added to 125 mL of a 60%
acid solution. How many mL of pure acid must be added to
obtain a solution of 83% acid? a) Find a function that finds the
concentration of the new mixture. b) Write and solve the
equation that answers the question.
x+0.6(125)
C(x)=
x+125
169.12
Finding a Minimum Perimeter


Find the least amount of fencing if the area must be
500 ft2.
Only 3 sides are needed. The 4th side is a building.
x
500 ft2
500
x
x
Finding a Minimum Perimeter
The perimeter is the function we want to
minimize.
500
f ( x)  2 x 
x
Finding a Minimum Perimeter
[-1, 40] x [-1, 300]

The sides should be 15.8 ft and 31.6 ft for a
perimeter of 63.25 ft.
Now You Try



Considering all rectangles with an area of
182 ft2. Let x be the length of one side of
such a rectangle.
Express the perimeter P as a function of x.
Find the dimensions of the rectangle that has
the least perimeter. What is the least
perimeter?
53.96 ft
Designing a Juice Can

Stewart Cannery will package tomato juice in
2-liter cylindrical cans. Find the radius and
height of the cans if the cans have a surface
area of 1000 cm2.
What We Need To Know
r = radius of can
 1 L = 1000 cm3
h = height of can
V = r2h

SA = 2r2 + 2rh
V
h 2
r
 2000 
1000  2 r  2 r 
2 
 r 
2
Solving the Equation
 2000 
1000  2 r 2  2 r 
2 

r


r=4.62
h = 29.63
[2, 10] x [700, 1100]
r=9.65
h = 6.83
With a surface area of 1000 cm2, the cans either
have a radius of 4.62 cm and a height of 29.83 cm
or have a radius of 9.65 and a height of 6.83 cm.
Now You Try



Drake Cannery will pack peaches in 0.5-L
cylindrical cans. Let x be the radius of the
can in cm.
Express the surface area S of the can as a
function of x.
Find the radius and height of the can if the
surface area is 900 cm2.
r = 1.12
h = 126.88
r = 11.37
h = 1.23
Home Work
253 – 256
#2, 8, 12, 14, 18, 32, 38,
45-50
P.